$\triangle DEF$ is inscribed inside $\triangle ABC$ such that $D,E,F$ lie on $BC, AC, AB$, respectively. The circumcircles of $\triangle DEC, \triangle BFD, \triangle AFE$ have centers $O_1,O_2,O_3$, respectively. Also, $AB = 23, BC = 25, AC=24$, and $\stackrel{\frown}{BF} = \stackrel{\frown}{EC},\ \stackrel{\frown}{AF} = \stackrel{\frown}{CD},\ \stackrel{\frown}{AE} = \stackrel{\frown}{BD}$. The length of $BD$ can be written in the form $\frac mn$, where $m$ and $n$ are relatively prime integers. Find $m+n$.

[asy] size(150); defaultpen(linewidth(0.8)); import markers; pair B = (0,0), C = (25,0), A = (578/50,19.8838); draw(A--B--C--cycle); label("$B$",B,SW); label("$C$",C,SE); label("$A$",A,N); pair D = (13,0), E = (11*A + 13*C)/24, F = (12*B + 11*A)/23; draw(D--E--F--cycle); label("$D$",D,dir(-90)); label("$E$",E,dir(0)); label("$F$",F,dir(180));  draw(A--E,StickIntervalMarker(1,3,size=6));draw(B--D,StickIntervalMarker(1,3,size=6)); draw(F--B,StickIntervalMarker(1,2,size=6)); draw(E--C,StickIntervalMarker(1,2,size=6)); draw(A--F,StickIntervalMarker(1,1,size=6)); draw(C--D,StickIntervalMarker(1,1,size=6));  label("24",A--C,5*dir(0)); label("25",B--C,5*dir(-90)); label("23",B--A,5*dir(180)); [/asy]
From adjacent sides, the following relationships can be derived:
\begin{align*} DC &= EC + 1\\ AE &= AF + 1\\ BD &= BF + 2 \end{align*}
Since $BF = EC$, and $DC = BF + 1$, $BD = DC + 1$. Thus, $BC = BD + DC = BD + (BD - 1)$. $26 = 2BD$. Thus, $BD = 13/1$. Thus, the answer is $\boxed{14}$.